Amp Meter

Hi guys
I have a circuit with a 24 volt DC power supply 2 amps that feeds about 20
small relays and I have an amp meter that just says 1 to 5 on it. How would
I put this in the circuit I think it would go in series but to draw 2 amps
through the meter would that not damage it
Thanks
Nikki

 

most likely the meter is already designed to be used inline directly.
other wise you would need to know the R of the meter coil and voltage
of the meter to calculate for a shunt resistor.

 

The meter itself should tell you, if the face is still legible. Since it's
a 1-5 amp meter, it should be OK, but if it has fine print near the bottom
of the face, read that, and if that doesn't tell you exactly how to
connect it, then post what it says, and we'll try to answer better.

Just in case I'm missing something, it _does_ have a 0 to the left of
the 1, doesn't it?

If so, and if there's nothing in fine print, and nothing stamped or molded
into the back of the meter, and if the studs are pretty good-sized, (like
5 mm dia. or fatter), then it's probably OK to just put it in series.
First thing out of the supply, of course. ;-)

Good Luck!
Rich

 

On closer inspection of the meter I have taken the cover off and the needle
will not move so I think the meter is no good. I have a second I would like
to use this one goes from 0 to 60 On the back it has a + on one of the posts
and printed on the back is 120 on the front of the meter is says (its very
small print) FS-50mvDC and it also says D C Amperes. Now I know this came
from the Phone company and it was on a 48 volt circuit.
Nikki

 

It sounds like the meter mechanism inside is a 50mV full scale
instrument, but it is very likely that it has a series or parallell
resistor inside the cover to make it suitable for 48 Volt or a few
amperes.

Open the cover and try to find a series or parallell resistor. Or simply
test the meter very carefully, using a battery and a high value resistor
in series, to limit the current to less than 10mA to begin with. If there
is no reaction change to a ten times lower resistor until it works.

 

Amp meters have a sensitive movement connected in parallel with a low
resistance current carrying shunt that produces the small voltage
required to move the meter (50 mV in this case). The open question
with any amp meter is whether that low resistance shunt is sealed
inside the meter or is assumed to be external to the meter. Measure
the resistance of your amp meter. If it is less than an ohm, the
shunt is almost certainly inside the meter, and it is safe to pass
amperes through the meter terminals. If higher than an ohm, the meter
is supposed to have an external shunt resistor connected to it, and
the load current is supposed to be passed through that resistor
through an extra pair of terminals. Here is what an external shunt
resistor might look like:
http://www.carrel.co.nz/shunt/main.htm
Note the big terminals for the load current and the smaller ones for
the meter connection.

A more detailed review of ammeters:
http://www.allaboutcircuits.com/vol_1/chpt_8/4.html

 

DO NOT DO THIS WITH AN OHMMETER!!!

Use two potentiometers.

Start with one in series with the meter and, say, a 1.5V battery. Start
with the pot at max. resistance. Turn it down until the meter reads
full-scale. Start with at least 10K. If a 1.5V battery won't give full-
scale deflection even with the pot turned to 0, then clearly the meter
expects the full 60A through itself, and you'd stick it in series with
the supply. Otherwise, go to step 2:

2: Take the second pot and put it in parallel with the meter (while
leaving the pot and battery, that have your meter deflected full-scale).
Adjust the second pot until the meter deflects exactly one-half. Take that
resistor, and measure _its_ resistance with the ohmmeter, and that will
tell you the internal resistance of the meter.

Then, it's just Ohm's law.

Good Luck!
Rich

 

you will need to determine the coil R (resistance), 120 maybe it.
lets assume for the moment that it is.
your scale is 0..60, we will use that for amp scale.

Imeter = 0.050/120 = 416 uA ( does not seem likely etc).

Rshunt = 0.050/(60.0 - 0.000416) = 0.00083 ohms

so this would mean you need a shunt resistor on the
meter terminals of 0.00083 which is more like a bar
of copper across the terminals of the meter.
if the 120 number you are reading is actually indicating
I of the meter then you use 120 uA or 120 mA instead of the
416 uA in the first calculation.

in either case, you will need to refer to a wire&meteral
chart to get teh ohms per 1K' (per 1k feet) to determine what
you need to create a shunt .

these are only my thoughts off the top of my head going by the
info your passing.

 

Guys thanks for all your replies
I took the suggestion of Rich and found that with a 1.5volt battery and a
resistance of 10400 ohms I can get the meter to read at its last increment.
The meter is marked in increments of 10 up to 60 how would I get a 2 amp
draw to read 20 where a 6 amp draw would be the last increment on the meter
Thanks
Nikki

 

Let's imagine a circuit in our minds. The instrument is in parallell with
a big shunt, we are seeing it deflect to the full 60 Amp, as you wanted
it.

What do we know about this circuit?

The instrument is a 65mV 148uA full scale instrument.
(1.5V/10400 Ohm = 148uA, and wasn't it a 65mV instrument?)

The voltage over the shunt must be 65mV too.

This shunt must have a resistance of 65mV divided by 20 Amps, which is
0.00325 Ohm.

So you need to find a piece of metal with this resistance from end to
end, then connect your instrument in parallell to it, and you have a 60
Amp current meter.

Make it a little longer than needed, and adjust the connection points of
the meter to the shunt to adjust it. Connect the outer current at the
real ends of the shunt.

That would create an instrument which does not need a power supply, no
battery.

If we add a battery we can choose a shunt much more freely, as we can use
an opamp to scale the result to fit to the instrument. It will give
adjustment possibilities and it is a lot easier to find a shunt.

 

I see now that you wanted a full scale of 6 Amp.
Well, adjust the calculations accordingly.

 

There is also a practical way to solve this problem without calculations.

Try different pieces of metal, pieces of rebar, metal bars of all kinds.

Send a known current through it. Put the probes from the instrument
together at the middle of the metal bar. Pull the probe tips apart slowly
and watch how the instrument shows higher indications. When at the full
scale, or the number you want, mark the two points.

Cut off the metal bar right outside the marks.
Connect the probes at their marked points, and the outer current from the
outer ends of the metal bar.

 

---------------
You will have to do a bit better. If you have a voltmeter, measure the
voltage at the battery- it is unlikely to be 1.5V. You could also measure
the voltage across the potentiometer and determine the actual current. It
appears to be a 50mv meter which seems to be reading full scale at
120ma.giving a resistance of 0.417 ohms.
If this is so -then for 6A full scale, the meter will need an external shunt
of 0.0085 ohms. You will need a wire large enough to carry 6A and long
enough to get 0.0085 ohms (4 inches of #24 wire may do).
However- check the actual current and voltage as above as the FS reading may
not be 120ma.
I assume you are not looking for great accuracy.
See calculations below
V=meter FS voltage. Im=meter FS current. I=desired full scale current
Rshunt =V/(I-Im)
In your case V=50mv, I=6A and, if the assumption above is correct, Im
=0.120A
then Rshunt =0.05/(6-0.12) =0.0085 ohms
#24 wire should carry 6A and has a resistance at 20degrees C of 25.7
ohms/1000ft so 4 inches should do.

Someone please check the above.

 

Thanks guys
If I have a circuit lets say 1.5 volt bat and a 1.5 volt lamp and I want to
measure the current I would put the meter parallel to a shunt resistor in
the circuit right. But does the shunt resistor not cut down on the voltage
to the lamp

 

I think you are right.

In practical terms: Take a piece of copper wire, diameter 1-2mm.
Send 1 Amp through it. Put the wires from the meter together on that
wire. Pull them slowly apart until the meter reads 1 Amp. Mark these
measuring points and solder the instrument wires to them. Cut off the
thick wire outside the measuring marks and attach the current connections
to the ends of the wire.

 

I've also seen a circuit where a microammeter is turned into a voltmeter -
in this case, 10.4K gives 1.5V full-scale. A shunt resistor that drops
1.5V at 60A would be .025 ohm, which is much more manageable of a value
than 0.0083 ohm, and the calibration isn't anywhere near as persnickety.

Good Luck!
Rich

 

Yes, by 0.05 volts.

But don't just slap a shunt in there and expect to get accurate readings.
Either give yourself a way of trimming it while live, or actually
determine the internal resistance of the meter. Take your setup with
the batt. and series R, adjust it for full-scale, as you've done, and
then take another pot, put it in parallel with the meter (so the meter
and new pot are now in parallel with each other, and that combination
is in series with the B and R), and adjust it until the meter reads
half-scale.

But this will only work if the meter actually _has_ a Zero indication.

Did you meant to say that your meter face has "1" at the very leftmost
spot? If so, you might have to do some more experiments to find out
exactly what this meter is expecting - usually, when there's a meter
that doesn't start at 0, it's used with some kind of custom circuit to
get that offset.

Of course, the other thing (depending on how much headroom you have)
is to turn the meter into a voltmeter, by just adding a series resistor,
and measuring the voltage across the shunt. Doing that gets your
calibration up out of the mud.

Good Luck!
Rich

 

In this case you are wasting a major part of the signal. That is not a
good idea, seen from an engineering point of view.

We want a shunt which gives as low voltage drop as possible. Your
idea would make the voltage drop many times bigger, from 50mV to 1500mV.

The 0.0083 Ohm shunt is easy to construct from heavy gauge copper wire.
Or a thinner wire laid double, in quadruple, etc..

Adjustment can be made by moving the probe measuring point a little, or
by connecting an adjustment pot in parallell with the shunt. A wirewound
pot will be suitable for this.

 

that looks to be a 150 uA meter at full scale.
if memory serves,.i think you said the meter was a 0.050 mv?

to get a 6 amp scale.
6.00 - .000150 = 5.999850

Shunt = 0.050 / 5.999850 = 0.0083 ohms

i think that is correct,i normally measure the
coil R and use the I (current) to do all of this
how ever, this should still work.
this is just off the top of my head, i am sure some
one will correct me.

 

Well, in my defense, in a different post, I did mention that this
technique depends on the headroom of the supply. And you can certainly
decrease the ranging resistors, to decrease the shunt drop. It's just
a seat-of-the-pants sort of thing - you decrease the pickiness of the
shunt by adding _some_ resistance in series with the meter.

I've calibrated ammeter shunts, and it's a PITA. They wouldn't let
me grind on the brass strip with the Dremel while power was on. ;-)

I've also seen ammeters that have no leads at all - they have a clip
on the back that clips over the cable, and the meter itself is nothing
but a spring-loaded compass. You calibrate them by moving the cable
around until the meter reads right, then hot-glue it.

Cheers!
Rich